MATH 225 Summer 2005 Linear Algebra II Solutions to Assignment 1 Due: Wednesday July 13, 2005
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1 MATH 225 Summer 25 Linear Algebra II Solutions to Assignment 1 Due: Wednesday July 13, 25 Department of Mathematical and Statistical Sciences University of Alberta Question 1. [p 224. #2] The set of all continuous real-valued functions defined on a closed interval [a, b] in R is denoted by C[a, b]. (a) Show that C[a, b] is a subspace of the vector space of all real-valued functions defined on [a, b]. (b) Let V = {f C[a, b] : f(a) = f(b)}. Show that V is a subspace of C[a, b]. (a) From your elementary calculus course, you know the following (i) The constant function f(t) =, a t b, is continuous on the interval [a, b]. (ii) If f and g are two continuous functions on the interval [a, b], then the function f + g defined by (f + g)(t) = f(t) + g(t), a t b, is also continuous on the interval [a, b]. (iii) If f is a continuous function on [a, b] and λ R is a scalar, then the function λf defined by (λf)(t) = λf(t), a t b, is also continuous on the interval [a, b]. Since C[a, b] contains the zero vector, is closed under addition and scalar multiplication, then it is a subspace of the vector space of all real-valued functions defined on [a, b]. (b) Let H = {f C[a, b] : f(a) = f(b)}. (i) If f(t) = for all t [a, b], then f(a) = f(b) =, and f H. (ii) If f, g C[a, b], then (f + g)(a) = f(a) + g(a) = f(b) + g(b) = (f + g)(b), and f + g H. (iii) Iff H and λ R, then (λf)(a) = λf(a) = λf(b) = (λf)(b), and λf H. Since H contains the zero vector, is closed under addition and scalar multiplication, then it is a subspace of the vector space C[a, b]. Question 2. [p 224. #22] Let F be a fixed 3 2 matrix, and let H be the set of all matrices A in M 2 4 with the property that F A = (the zero matrix in M 3 4 ). Determine if H is a subspace of M 2 4. (i) If A is the M 2 4 zero matrix, then F =, and H contains the zero vector. (ii) If A and B are in H, then F (A + B) = F A + F B = + =, and A + B H. (iii) If A is in H and λ R, then F (λa) = λf A = λ =, and λa H. Therefore H contains the zero vector and is closed under addition and scalar multiplication, so H is a subspace of M 2 4. Question 3. [p 225. #32] Let H and K be subspaces of a vector space V. The intersection of H and K, written as H K, is the set of v in V that belong to both H and K, that is, H K = {v V : v H and v K}. Show that H K is a subspace of V. Give an example in R 2 to show that the union of two subspaces is not, in general, a subspace.
2 Let H and K be subspaces of the vector space V, then (i) Since H and K both contain, then H K. (ii) If u, v H K, then since H and K are closed under addition, we have u + v H K. (iii) If v H K and λ R, then since H K is closed under scalar multiplication, we have λv H K. Therefore H K contains the zero vector and is closed under addition and scalar multiplication, so H K is a subspace of V. Let H = span{(1, )} and K = span{(, 1)}, then H K = {(u, ) : u R} {(, v) : v R}, but (1, ) + (, 1) = (1, 1) H K, and the union of these two subspaces is not closed under addition, so is not a subspace of R 2. Question 4. [p 235. #3] Let T : V W be a linear transformation from a vector space V into a vector space W. Prove that the range of T is a subspace of W. [Hint: Typical elements of the range have the form T (x) and T (w) for some x, w in V.] Let T (V ) = {b W : b = T (v) for some v V } be the range of T. (i) Since = T () and V, then T (V ). (ii) Let u, v T (V ), then there exist x, y V such that u = T (x) and v = T (y), and therefore we have u + v = T (x) + T (y) = T (x + y) T (V ) since T is additive and V is closed under addition. (iii) Let v T (V ) and let λ R, then v = T (x) for some x V, so that λv = λt (x) = T (λv) T (V ) since T is homogeneous and T (V ) is closed under scalar multiplication. Therefore the range of T contains the zero vector of W, is closed under addition and scalar multiplication, so T (V ) is a subspace of W. Question 5. [p 236. #34] Define T : C[, 1] C[, 1] as follows: For f in C[, 1], let T (f) be the antiderivative F of f such that F() =. Show that T is a linear transformation and describe the kernel of T. If f C[, 1], from the fundamental theorem of calculus, the antiderivative of f that vanishes at is given by for x 1. If f, g C[, 1], then T (f + g)(x) = T (f)(x) = F (x) = (f(t) + g(t)) dt = f(t) dt + f(t) dt for all x [, 1], so that T (f + g) = T (f) + T (g), and T is additive. If f C[, 1] and λ R, then T (λf)(x) = λf(t) dt = λ 1 g(t) dt = T (f)(x) + T (g)(x) f(t) dt = λt (f)(x) for all x [, 1], so that T (λf) = λt (f), and T is homogeneous. Therefore T is linear.
3 Note that f ker(t ) if and only if T (f) =, that is, T (f)(x) = f(t) dt = for all x [, 1], that is, the constant function whose value is for all x [, 1]. Again, from the fundamental theorem of calculus, if for all x [, 1], then since f is continuous we have for all x [, 1]. Therefore, ker(t ) = {}. Question 6. [p 236. #36] f(t) dt = f(x) = d ( ) f(t) dt = dx Let V and W be vector spaces and let T : V W be a linear transformation. Let Z be a subspace of W, and let U be the set of all x in V such that T (x) is in Z. Show that U is a subspace of V. This subspace is usually denoted by T 1 (Z). (i) Since Z is a subspace of W, then Z, and since T is linear, then T () =, so that T 1 (Z). (ii) If x, y T 1 (Z), then T (x) Z and T (y) Z, since Z is a subspace then T (x + y) = T (x) + T (y) Z, and therefore x + y T 1 (Z). (iii) If x T 1 (Z) and λ R, then T (x) Z, and since T is a linear transformation and Z is a subspace, then T (λx) = λt (x) Z, so that λx T 1 (Z). Since T 1 (Z) and T 1 (Z) is closed under addition and scalar multiplication, then T 1 (Z) is a subspace of V. Question 7. [p 244. #24] Let B = {v 1, v 2,..., v n } be a linearly independent set in R n. Explain why B must be a basis for R n. Let A = (v 1 v 2 v n ) be the n n matrix whose columns are the vectors in B and let U be the reduced row echelon form of A. Since A is a square matrix and Ax = if and only if Ux =, then the n columns of U are linearly independent, so that U = I, the n n identity matrix. Therefore, by the Gaussian elimination algorithm, for each b R n, the equation Ax = b has a unique solution, that is, b = x 1 v 1 + x 2 v x n v n for some scalars x 1, x 2,..., x n, so that the columns of A span R n also. Therefore, B = {v 1, v 2,..., v n } is a basis for R n. Question 8. [p 244. #29] Let S = {v 1,..., v k } be a set of k vectors in R n, with k < n. Explain why S cannot be a basis for R n. Let A = (v 1 v 2 v k ) be the n k matrix whose columns are the vectors in S and let U be the reduced row echelon form of A, since k < n, then U has fewer columns than rows, and therefore does not have n leading 1 s, so that U has at least one row consisting of all s. This means that for some vectors b R n, the system Ax = b will be inconsistent, so it is not true that Ax = b has a solution for each b R n, that is, the columns of A do not span R n, and so do not form a basis for R n.
4 Question 9. [p 244. #3] Let S = {v 1,..., v k } be a set of k vectors in R n, with k > n. Explain why S cannot be a basis for R n. Let A = (v 1 v 2 v k ) be the n k matrix whose columns are the vectors in S and let U be the reduced row echelon form of A, since k > n, then U has fewer rows than columns, and therefore does not have k leading 1 s, so that the system Ux = must have nontrivial solutions, so the columns of U are linearly dependent. However, since Ax = and Ux = have exactly the same solutions, then the corresponding set of columns in A is also linearly dependent, and so S cannot be a basis for R n. Question 1. [p 254. #18] Let B = {b 1,..., b n } be a basis for a vector space V. Explain why the B-coordinate vectors of b 1,..., b n are the columns e 1,..., e n of the n n identity matrix. For each of the basis vectors, say b k, we have b k = b b k b k + b k b n so that for k = 1, 2,..., n. [b k ] B = (,...,, 1,,..., ) = e k Question 11. [p 254. #24] Let V be a vector space with basis B = {b 1,..., b n } and the coordinate mapping x [x] B. Show that the coordinate map is onto R n, that is, given any y in R n, with entries y 1,..., y n, produce a u in V such that [u] B = y. Let y = (y 1, y 2,..., y n ) R n, and let u = y 1 b 1 + y 2 b y n b n, then [u] B = y, and since y R n was arbitrary, then the coordinate mapping x [x] B is onto R n. Question 12. [p 254. #25] Let V be a vector space with basis B = {b 1,..., b n } and the coordinate mapping x [x] B. Show that a subset {u 1,..., u p } in V is linearly independent if and only if the set of coordinate vectors {[u 1 ] B,..., [u p ] B } is linearly independent in R n. Hint: Explain why, since the coordinate mapping is one-to-one, the following equations have the same solutions, c 1,..., c p. c 1 u c p u p = (zero vector in V ) c 1 [u 1 ] B + + c p [u p ] B = [] B (zero vector in R n ) Since the coordinate mapping is one-to-one, then the equations and c 1 u c p u p = (zero vector in V ) [c 1 u c p u p ] B = (zero vector in R n ) have exactly the same solutions, and since the coordinate mapping is linear, these equations have exactly the same solutions as c 1 [u 1 ] B + + c p [u p ] B = [] B (zero vector in R n ).
5 Therefore, {u 1,..., u p } is a linearly independent set in V if and only if {[u 1 ] B,..., [u p ] B } is a linearly independent set in R n. Question 13. [p 261 #21] The first four Hermite polynomials are 1, 2t, 2 + 4t 2, and 12t + 8t 3. These polynomials arise naturally in the study of certain important differential equations in mathematical physics. Show that the first four Hermite polynomials form a basis for P 3. The matrix whose columns are the coordinate vectors of the Hermite polynomials relative to the standard basis {1, t, t 2, t 3 } of P 3 is given by A =. 4 8 Since this matrix is already in row echelon form and there are 4 nonzero pivots, then its columns are linearly independent. Since the coordinate vectors form a linearly independent set, then the Hermite polynomials form a linearly independent set in P 3. The dimension of P 3 is 4, so this set of Hermite polynomials forms a basis for P 3. Question 14. [p 261. #22] The first four Laguerre polynomials are 1, 1 t, 2 4t + t 2, and 6 18t + 9t 2 t 3. Show that these polynomials form a basis for P 3. The matrix whose columns are the coordinate vectors of the Laguerre polynomials relative to the standard basis {1, t, t 2, t 3 } of P 3 is given by A = Since this matrix is already in row echelon form and there are 4 nonzero pivots, then its columns are linearly independent. Since the coordinate vectors form a linearly independent set, then the Laguerre polynomials form a linearly independent set in P 3. The dimension of P 3 is 4, so this set of Laguerre polynomials forms a basis for P 3. Question 15. [p 262. #28] Show that the space C(R) of all continuous functions defined on the real line is an infinite-dimensional vector space. Suppose that C(R) is finite dimensional, then this vector space has a basis consisting of m vectors, where m is a positive integer. Now note that the set of polynomials {1, t, t 2,..., t m 1, t m } is a linearly independent set in C(R), using the fact that C(R) can be spanned by a set of m vectors, this implies that m + 1 m, which is a contradiction. Therefore, C(R) is infinite dimensional.
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